This lesson aims at showing how to get the following physical properties :
This first lesson covers the sections 1, 3, 4 and 6 of the abinit_help file.
The very first step is a detailed tour of the input and output files : you
are like a tourist, and you discover a town in a coach. You will have a bit
more freedom after that first step ...
It is supposed that you have some good knowledge of UNIX/Linux.
This lesson should take about 2 hours.
1.1.a In addition to the present window, open the second window. Go to the Tutorial directory (that we refer as ~abinit/tests/tutorial/Input).
1.1.b You also need a working directory. So, you should create a subdirectory of this directory, whose name might be "Work" (so ~abinit/tests/tutorial/Input/Work). Change the working directory of windows 2 to "Work":
You will do most of the actions of this tutorial in this working directory. Copy the file tbase1_x.files in "Work" :
1.1.c Edit the tbase1_x.files. It is not very long (only 6 lines). It gives the information needed for the code to build other file names ... You will discover more about this file in the section 1.1 of the abinit_help file. Please, read it now (it will take one minute or so).
1.1.d Modify the first and second lines of tbase1_x.files file, so that they read:
Later, you will again modify these lines, to treat more cases. You should also change the last line, that gives the location of the pseudopotential (suppress one occurrence of "../"). Close the tbase1_x.files file. Then, copy the file ~abinit/tests/tutorial/Input/tbase1_1.in in "Work" :
Also later, we will look at this file, and learn about its content. For now, you will try to run the code. Its name is 'abinit'. The place where it can be found varies, according to the installation procedure. We will denote the directory where you have installed the package ~abinit. Supposing that you dumped the binaries from the Web site, then 'abinit' is to be found in the package, with location ~abinit/opt . If you dumped the sources from the Web site, and issued ./configure in the ~abinit directory, then it is located in ~abinit/src/main . In what follows, we will suppose that you can call it by simply typing "abinit", even if the actual command must be something like ../../../../opt/abinit or ../../../../src/main/abinit . (Suggestion : use an alias, copy the abinit executable, or declare the path).
So, in the Work directory, type :
abinit log tbase1_1.in tbase1_1.out tbase1_x.files tbase1_xo_DDB tbase1_xo_DEN tbase1_xo_EIG tbase1_xo_WFKDifferent output files have been created, including a "log" file and the output file "tbase1_1.out". To check that everything is correct, you can make a diff of tbase1_1.out with a reference file contained in the ~abinit/tests/tutorial/Refs directory :
(Perhaps you will need to ignore the blanks, with the command "diff -b" instead of "diff")
That reference file uses slightly different file names. You should get some difference, but rather inoffensive ones, like differences in the name of input files or timing differences, e.g. :
2,3c2,3 < .Version 5.5.4 of ABINIT < .(sequential version, prepared for a powerpc_darwin8.6.0_gfortran4.2 computer) --- > .Version 5.5.3 of ABINIT > .(sequential version, prepared for a x86_64_linux_intel9.1 computer) 17c17 < .Starting date : Sat 24 May 2008. --- > .Starting date : Fri 29 Feb 2008. 27c27 < - input file -> tbase1_1.in --- > - input file -> ../tbase1_1.in 29,30c29,30 < - root for input files -> tbase1_xi < - root for output files -> tbase1_xo --- > - root for input files -> tbase1_1i > - root for output files -> tbase1_1o 92,93c92,93 < - pspini: atom type 1 psp file is ../../../Psps_for_tests/01h.pspgth < - pspatm: opening atomic psp file ../../../Psps_for_tests/01h.pspgth --- > - pspini: atom type 1 psp file is /home/gonze/ABINIT/ABINITv5.5.3/trunk/5.5.3-private/tests/Psps_for_tests/01h.pspgth > - pspatm: opening atomic psp file /home/gonze/ABINIT/ABINITv5.5.3/trunk/5.5.3-private/tests/Psps_for_tests/01h.pspgth 166c166 < prteigrs : about to open file tbase1_xo_EIG --- > prteigrs : about to open file tbase1_1o_EIG 214c214 < - Total cpu time (s,m,h): 4.7 0.08 0.001 --- > - Total cpu time (s,m,h): 4.6 0.08 0.001 221,229c221,228(... and what comes after that is related only to timing ...). If you do not run on a PC under Linux with Intel Fortran compiler, you might also have small numerical differences, on the order of 1.0d-10 at most. You might also have other differences in the paths of files. If you get something else, you should ask for help !
Supposing everything went well, we will now detail the different steps that took place : how to run the code, what is in the "tbase1_1.in" input file, and, later, what is in the "tbase1_1.out" and "log" output files.
1.1.e Running the code is described in the section 1.2 of the abinit_help file. Please, read it now (it will take 30 seconds or less).
1.1.f It is now time to edit the tbase1_1.in file. You can have a first glance at it. It is not very long: about 50 lines, mostly comments. Do not try to understand everything immediately. After having gone through it, you should read general explanation about its content, and the format of such input files in the section 3.1 of the abinit_help file.
1.1.g You might now examine in more details some input variables. An alphabetically ordered index of all variables is provided, and their description is found in the following files:
It is now time to have a look at the two output files of the run.
1.1.h First, edit the "log" file. You can begin to read it. It is nasty. Jump to its end. You will find there the number of WARNINGS and COMMENTS that were issued by the code during execution. You might try to find them in the file (localize the keywords "WARNING" or "COMMENT" in this file). Some of them are for the experienced user. For the present time, we will ignore them. You can find more information about messages in the log file in the section 6.1 of the abinit_help file.
1.1.i Then, edit the "tbase1_1.out" file. You find some general information about the output file in section 6.2 of the abinit_help file. You should also :
If the code does not stop there, the input parameters are consistent. At this stage, many default values have been provided, and the preprocessing is finished.
It is worth to come back to the echo of preprocessed input data. You should
first examine the "tbase1_1.in" file in more
details, and read the meaning of each of its variables in the
corresponding input variables
file, if it has not yet been done. Then, you should examine some variables
that were NOT defined in the input file, but that appear in the echo written
in "tbase1_1.out" :
- "nband" : its value is 2.
It is the number of electronic states that will be treated by the code. It has been computed by counting the number of valence electrons in the unit cell (summing the valence electrons brought by each pseudopotential) then occupying the lowest states (look at the "occ" variable), and adding some states (at least one, maybe more, depending on the size of the system).
- "ngfft" : its value is 30 30 30 .
It is the number of points of the three-dimensional FFT grid. It has been derived from "ecut" and the dimension of the cell ("acell").
The maximal number of plane waves "mpw" is mentioned in the memory evaluation section: it is 752.
Well, this is not completely right, as the code took advantage of the time-reversal symmetry, valid for the k-point (0 0 0), to decrease the number of planewave by about a factor of two.
The full set of plane waves is 1503 (see later in the "tbase1_1.out" file).
The code indicates the time-reversal symmetry by a value of istwfk=2 , instead of the usual istwfk=1 default.
- "nsym" : its value is 16.
It is the number of symmetries of the system. The 3x3 matrices symrel define the symmetries operation. In this case, none of the symmetries is accompanied by a translation, that would appear in the variable "tnons". The code did an automatic analysis of symmetries.
They could alternatively be set by hand, or using the symmetry builder (to be described later).
- "xangst" and "xred" are alternative ways to "xcart" to specify the positions of atoms within the primitive cell.
Now, you can start reading the description of the remaining of the tbase1_1.out file, in the section 6.3 of the abinit_help file. Look at the tbase1_1.out file at the same time.
1.1.j You have read completely an output file !
Could you answer the following questions ?
(answers are given at the end of the present file)
1.2.a Starting from now, everytime a new input variable is mentioned, you should read the corresponding descriptive section in the ABINIT help.
We will now complete the description of the meaning of each term : there are still a few indications that you should be aware of, even if you will not use them in the tutorial. These might appear in the description of some input variables ... For this, you should read the section 3.2 of the abinit_help file.
1.2.b There are three methodologies to compute the optimal distance between the two Hydrogen atoms:
The interatomic distance in the tbase1_1.in file was 1.4 Bohr. Suppose you decide
to examine the interatomic distances from 1.0 Bohr to 2.0 Bohr, by steps of
0.05 Bohr. That is, 21 calculations.
If you are a UNIX guru, it will be easy for you to write a script that will drive these 21 calculations, changing automatically the variable "xcart" in the input file, and then gather all the data, in a convenient form to be plotted.
Well, are you a UNIX guru ? If not, there is an easier path, all within ABINIT !
This is the multi-dataset mode. Detailed explanations about it can be found in sections 3.3, 3.4, 3.5 and 3.6, of the abinit_help file.
1.2.c Now, can you write an input file that will do the computation described above (interatomic distances from 1.0 Bohr to 2.0 Bohr, by steps of 0.05 Bohr) ? You might start from tbase1_1.in . Try to define a series, and to use the "getwfk" input variable (the latter will make the computation much faster).
You should likely have a look at the section that describes the "irdwfk" and "getwfk" input variables: in particular, look at the meaning of getwfk -1
Also, define explicitely the number of states (or supercell "bands") to be one, using the input variables "nband". The input file ~abinit/tests/tutorial/Input/tbase1_2.in is an example of file that will do the job, while ~abinit/tests/tutorial/Refs/tbase1_2.out is an example of output file. If you decide to use the ~abinit/tests/tutorial/Input/tbase1_2.in file, do not forget to change the file names in the tbase1_x.files file ...
So, you run the code with your input file (this might take fifteen seconds or so on a PC at 3 GHz), examine the output file quickly (there are many repetition of sections, for the different datasets), and get the output energies gathered in the final echo of variables :
etotal1 -1.0368223891E+00 etotal2 -1.0538645433E+00 etotal3 -1.0674504851E+00 etotal4 -1.0781904896E+00 etotal5 -1.0865814785E+00 etotal6 -1.0930286804E+00 etotal7 -1.0978628207E+00 etotal8 -1.1013539124E+00 etotal9 -1.1037224213E+00 etotal10 -1.1051483730E+00 etotal11 -1.1057788247E+00 etotal12 -1.1057340254E+00 etotal13 -1.1051125108E+00 etotal14 -1.1039953253E+00 etotal15 -1.1024495225E+00 etotal16 -1.1005310615E+00 etotal17 -1.0982871941E+00 etotal18 -1.0957584182E+00 etotal19 -1.0929800578E+00 etotal20 -1.0899835224E+00 etotal21 -1.0867972868E+00
You might try to plot these data. The minimum of energy in the above list is clearly between dataset 11 and 12, that is :
xcart11 -7.5000000000E-01 0.0000000000E+00 0.0000000000E+00 7.5000000000E-01 0.0000000000E+00 0.0000000000E+00 xcart12 -7.7500000000E-01 0.0000000000E+00 0.0000000000E+00 7.7500000000E-01 0.0000000000E+00 0.0000000000E+00
corresponding to a distance of H atoms between 1.5 Bohr and 1.55 Bohr. The forces vanish also between 1.5 Bohr and 1.55 Bohr :
fcart11 -5.4945071285E-03 0.0000000000E+00 0.0000000000E+00 5.4945071285E-03 0.0000000000E+00 0.0000000000E+00 fcart12 6.9603067838E-03 0.0000000000E+00 0.0000000000E+00 -6.9603067838E-03 0.0000000000E+00 0.0000000000E+00
From these two values, using a linear interpolation, one get the optimal value
of 1.522 Bohr .
Note that the number of SCF cycles drops from 6 to 5 when the wavefunctions are read from the previous dataset.
1.3.a The other methodology is based on an automatic computation of the minimum.
There are different algorithms to do that. See the input variable "ionmov", with values 2, 3 and 7. In the present case, with only one degree of freedom to be optimized, the best choice is ionmov 3 .
You have also to define the maximal number of timesteps for this optimization. Set the input variable "ntime" to 10, it will be largely enough. For the stopping criterion "tolmxf", use the reasonable value of 5.0d-4 Ha/Bohr. This defines the force treshold to consider that the geometry is converged. The code will stop if the residual forces are below that value before reaching "ntime".
It is also worth to change the stopping criterion for the SCF cycle, in order to be sure that the forces generated for each trial interatomic distance are sufficiently converged. Indeed, the value used for toldfe, namely 1.0d-6, might be sufficient for total energy calculations, but definitely not for the accurate computation of other properties. So, change "toldfe" in "toldff", and set the latter input variable to ten times smaller than "tolmxf". The input file ~abinit/tests/tutorial/Input/tbase1_3.in is an example of file that will do the job, while ~abinit/tests/tutorial/Refs/tbase1_3.out is an example of output file. If you decide to use these files, do not forget to change the file names in the tbase1_x.files" file ... So, you run the code with your input file (a few seconds), examine quietly this file (which is much smaller than the tbase1_2.out file), and get some significant output data gathered in the final echo of variables :
etotal -1.1058360644E+00 fcart 1.8270533893E-04 0.0000000000E+00 0.0000000000E+00 -1.8270533893E-04 0.0000000000E+00 0.0000000000E+00 ... xcart -7.6091015760E-01 0.0000000000E+00 0.0000000000E+00 7.6091015760E-01 0.0000000000E+00 0.0000000000E+00
According to these data (see xcart), the optimal interatomic distance is about 1.522 Bohr, in good agreement with the estimation of tbase1_2.out . If you have time (this is to be done at home), you might try to change the stopping criteria, and redo the calculation, to see the level of convergence of the interatomic distance.
Note that the final value of fcart in your run might differ slightly from the one shown above (less than one percent change). Such a fluctuation is quite often observed for a value converging to zero (remember, we ask the code to determine the equilibrium geometry, that is, forces should be zero) when the same computation is done on different platforms.
1.4.a We start from the optimized interatomic distance 1.522 Bohr, and make a run at fixed geometry. The input variable "prtden" must be set to 1. To understand correctly the content of the "prtden" description, it is worth to read a much more detailed description of the "files" file, in section 4 of the abinit_help file.
1.4.b The input file ~abinit/tests/tutorial/Input/tbase1_4.in is an example of input file for a run that will print a density. If you decide to use this file, do not forget to change the file names in tbase1_x.files. The run will take a few seconds.
The density will be output in the tbase1_xo_DEN file. Try to edit it ... No luck ! This file is unformatted, not written using the ASCII code. Even if you cannot read it, its description is provided in the abinit_help. It contains first a header, then the density numbers. The description of the header is presented in section 6.4 of the abinit_help file, while the body of the _DEN file is presented in section 6.5. It is the appropriate time to read also the description of the potential files and wavefunctions files, as these files contain the same header as the density file, see sections 6.6 and 6.7
1.4.c Such a density file can be read by ABINIT, to restart a calculation
(see the input variable iscf,
when its value is -2), but more usually, by an utility called "cut3d". This
utility is available in the ABINIT package.
You might try to use it now, to generate two-dimensional
cuts in the density, and visualize the charge density contours.
Read the corresponding Cut3D help file. Then, try to run cut3d to analyse tbase1_xo_DEN. You should first try to translate the unformatted density data to indexed formatted data, by using option 6 in the adequate menu. Save the indexed formatted data to file tbase1_xo_DEN_indexed. Then, edit this file, to have an idea of the content of the _DEN files.
For further treatment, you might choose to select another option than 6. In particular, if you have access to MATLAB, choose option 5. With minor modifications (set ngx=ngy=ngz to 30) you will be able to use the file dim.m present in ~abinit/doc/tutorial/lesson_base1 to visualize the 3-Dimensional isosurfaces. Another option might be to use the XCrysDen software, for which you need to use option 9.
The atomisation energy is the energy needed to separate a molecule
in its constituent atoms, each being neutral.
In the present case, one must compute first the total energy of an isolated hydrogen atom. The atomisation energy will be the difference between the total energy of H2 and twice the total energy of H .
There are some subtleties in the calculation of an isolated atom.
The input file ~abinit/tests/tutorial/Input/tbase1_5.in is an example of file that will do the job, while ~abinit/tests/tutorial/Refs/tbase1_5.out is an example of output file. If you decide to use the tbase1_5.in file, do not forget to change the file names in the tbase1_x.files file. The run lasts a few seconds.
You should read the output file, and note the tiny differences related with the spin-polarisation :
Eigenvalues (hartree) for nkpt= 1 k points, SPIN UP: kpt# 1, nband= 1, wtk= 1.00000, kpt= 0.0000 0.0000 0.0000 (reduced coord) -0.26414 Eigenvalues (hartree) for nkpt= 1 k points, SPIN DOWN: kpt# 1, nband= 1, wtk= 1.00000, kpt= 0.0000 0.0000 0.0000 (reduced coord) -0.11112
Spin up density [el/Bohr^3] , Maximum= 1.4053E-01 at reduced coord. 0.0000 0.0000 0.0000 ,Next maximum= 1.2019E-01 at reduced coord. 0.0000 0.0000 0.9667 , Minimum= 3.4544E-06 at reduced coord. 0.4667 0.4333 0.4333 ,Next minimum= 3.4544E-06 at reduced coord. 0.5333 0.4333 0.4333 Spin down density [el/Bohr^3] , Maximum= 0.0000E+00 at reduced coord. 0.9667 0.9667 0.9667 ,Next maximum= 0.0000E+00 at reduced coord. 0.9333 0.9667 0.9667 , Minimum= 0.0000E+00 at reduced coord. 0.0000 0.0000 0.0000 ,Next minimum= 0.0000E+00 at reduced coord. 0.0333 0.0000 0.0000 Magnetization (spin up - spin down) [el/Bohr^3] , Maximum= 1.4053E-01 at reduced coord. 0.0000 0.0000 0.0000 ,Next maximum= 1.2019E-01 at reduced coord. 0.0000 0.0000 0.9667 , Minimum= 3.4544E-06 at reduced coord. 0.4667 0.4333 0.4333 ,Next minimum= 3.4544E-06 at reduced coord. 0.5333 0.4333 0.4333 Relative magnetization (=zeta, between -1 and 1) , Maximum= 1.0000E+00 at reduced coord. 0.9667 0.9667 0.9667 ,Next maximum= 1.0000E+00 at reduced coord. 0.9333 0.9667 0.9667 , Minimum= 1.0000E+00 at reduced coord. 0.0000 0.0000 0.0000 ,Next minimum= 1.0000E+00 at reduced coord. 0.0333 0.0000 0.0000
The total energy is
etotal -4.7010531489E-01while the total energy of the H2 molecule is (see test 13):
The atomisation energy is thus 0.1656 Ha (The difference between the total energy of the H2 molecule and twice the energy of an isolated Hydrogen atom).
At this stage, we can compare our results :
What is wrong ??
Well, are you sure that the input parameters that we did not discuss are correct ? These are :
We will see in the next lesson how to address the choice of these parameters (except the pseudopotential).
Answers to the questions, section 1.1.j :
NOTE : there might be numerical differences, from platform to platform, in the quoted results !
Q1. 6 SCF cycles were needed :
iter Etot(hartree) deltaE(h) residm vres2 diffor maxfor ETOT 1 -1.1013391225242 -1.101E+00 4.220E-04 8.396E+00 2.458E-02 2.458E-02 ETOT 2 -1.1036939626391 -2.355E-03 7.374E-09 2.840E-01 1.325E-02 3.783E-02 ETOT 3 -1.1037170965209 -2.313E-05 7.389E-08 1.549E-02 1.207E-03 3.662E-02 ETOT 4 -1.1037223548790 -5.258E-06 4.146E-07 2.715E-04 8.561E-04 3.748E-02 ETOT 5 -1.1037224212232 -6.634E-08 4.091E-09 5.700E-06 7.091E-05 3.740E-02 ETOT 6 -1.1037224213136 -9.037E-11 5.808E-12 3.076E-07 1.238E-06 3.741E-02 At SCF step 6, etot is converged : for the second time, diff in etot= 9.037E-11 < toldfe= 1.000E-06Note that the number of steps that were allowed, nstep=10, is larger than the number of steps effectively needed to reach the stopping criterion. As a rule, you should always check that the number of steps that you allowed was sufficient to reach the target tolerance. You might now play a bit with nstep, as e.g. set it to 5, to see how ABINIT reacts.
Side note : in most of the tutorial examples, nstep will be enough to reach the target tolerance, defined by one of the "tolXXX" input variables. However, this is not always the case (e.g. the test case 1 of the lesson response-function 1), because of some portability problems, that could only be solved by stopping the SCF cycles before the required tolerance.
Q2. The information is contained in the same piece of the output file. Yes, the energy is more converged than toldfe, since the stopping criterion asked for the difference between successive evaluations of the energy to be smaller than toldfe twice in a row, while the evolution of the energy is nice, and always decreasing by smaller and smaller amounts.
Q3. These values are :
cartesian forces (hartree/bohr) at end: 1 -0.03740558871217 0.00000000000000 0.00000000000000 2 0.03740558871217 0.00000000000000 0.00000000000000 frms,max,avg= 2.1596127E-02 3.7405589E-02 0.000E+00 0.000E+00 0.000E+00 h/b
On the first atom (located at -0.7 0 0 in cartesian coordinates, in Bohr), the force vector is pointing in the minus x direction, and in the plus x direction for the second atom located at +0.7 0 0 .
The H2 molecule would like to expand ...
Q4. The eigenvalues (in Hartree) are mentioned at the lines
Eigenvalues (hartree) for nkpt= 1 k points: kpt# 1, nband= 2, wtk= 1.00000, kpt= 0.0000 0.0000 0.0000 (reduced coord) -0.36525 -0.01379
As mentioned in the abinit_help file, the absolute value of eigenenergies is not meaningful. Only differences of eigenenergies, as well as differences with the potential.
The difference is 0.35147 Hartree, that is 9.564 eV .
Moreover, remember that Kohn-Sham eigenenergies are formally NOT connected to experimental excitation energies !
(Well, more is to be said later about this ...).
Q5. The maximum electronic density in electron per Bohr cube is reached at the mid-point between the two H atoms :
Total charge density [el/Bohr^3] , Maximum= 2.6907E-01 at reduced coord. 0.0000 0.0000 0.0000